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I try to find an closed form answer the following integral, $$\int_{0}^{\infty}dx\,x^{m}J_l(Qx)e^{-x^2},\quad m\in\mathbb{Z}, l\in\mathbb{Z},Q>0$$ where $J_l(Qx)$ is Bessel function of the first kind, but I am not sure that it exists. Can anybody help me?

Wolfram Mathematica says that the answer (with assumption $\mathrm{Re}\,(l+m)>-1$) is $$2^{-l-1}Q^l\Gamma\left(\frac{m+l+1}{2}\right){}_1\tilde{F}_1\left(\frac{m+l+1}{2},l+1,-\frac{Q^2}{4}\right),$$ where ${}_1\tilde{F}_1$ is regularized confluent hypergeometric function. However, I feel that due to $m,l\in\mathbb{Z}$ more simple answer can exist.

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  • $\begingroup$ please define $J_l$ $\endgroup$ Mar 19 '20 at 20:25
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    $\begingroup$ @phdmba7of12 I have edited the question $\endgroup$ Mar 19 '20 at 20:29
  • $\begingroup$ not sure if this relationship would help at all (found on another post here) $$e^{-\frac{x^2}{4}}=\lim_{\nu\rightarrow\infty}\Gamma(\nu+1)\left(\frac{2}{\sqrt\nu x}\right)^\nu J_\nu\left(\sqrt{\nu} x\right),$$ $\endgroup$ Mar 19 '20 at 20:35
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    $\begingroup$ @phdmba7of12 to be honest, I do not understand how it can help. You suggest to plug this representation of exp into the integral? $\endgroup$ Mar 19 '20 at 20:36
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    $\begingroup$ It can be found in Gradshteyn and Ryzhik as formula number 6.631.1. $\endgroup$ Mar 19 '20 at 20:47

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